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For each ol the following graphs draw planar representation O1 show (hatt it has subgraph homeomorphic to K5 Kss...

Question

For each ol the following graphs draw planar representation O1 show (hatt it has subgraph homeomorphic to K5 Kss

For each ol the following graphs draw planar representation O1 show (hatt it has subgraph homeomorphic to K5 Kss



Answers

How many edges must be removed from a connected graph with n vertices and m edges to produce a spanning tree?

All right first let's suppose end plus five is even and we'll show that three and plus one is even if N plus five is even, then N has to be ought. Therefore N has to equal to K plus one. For some integer K. Then you can compute three N plus one equals six K plus three plus one, which equals six K plus four. And simplifies to two times three K plus two. In other words it is two times in integer. That is why at three N plus one must be even. It equals two times the integer three K plus two. For the other direction three N plus one is even implies and plus five is even well actually show that by contra positive, we'll assume and plus five is odd. And showed that three and plus one is odd. Well, if and plus fine design then and is even so N is two times some integer K. So three N plus one equals six K plus one, which equals two times the integer three K plus one. Therefore three M plus one is odd. Now for part B. One of the key facts you need is that a vertex in the complete graph on advertises has degree and minus one. So you look at the complete graph on three N plus two virgins is K. Three N plus two. That graph has an oiler circuit. If and only if all those degrees of the verses which are all the same. It's going to have to be even so three N plus one. The degree of all those virtues is that's going to have to be even by part A. We've proven that three N plus one is even if not only if N plus five is even and what's and close fine. It's the degree of all the vertex is in cayenne plus six. And if that's even if N plus five is even, then that means that complete graph can plus six has an oiler circuit. Therefore, the chain of equivalence is uh yields that can K. Three and plus two has an oiler circuit. If and only if KM plus six has an oiler, Sir.

So we need to draw the graph representing the following agency metrics. So let's write the metrics first. So we have 0002 Than 41 01 Then we have 11 02 Then we have 11 little too. So this is four by four matrix. Now let's name, this is A B C. D. Points. Similarly A B C and D. Points. So let's mark that points here. So we have A B. See and the point. Now let's check the relation. So for eight way there is no agency be no agencies. Senior agency A. Two D. We have to so a two D. Here we have to and similarly D. To A. We have one 32. A. So if this is A two D. This is G. Two and which is one. Now we'll move to the B. One so B to A. Is four. So be to A. Is for and similarly A. To B. That is already being checked. So between our B two, B is one. To be to be is one. Then we have B. Two C. Nor emergency and be two days one so B to D. Just one now see to A. Is one soc two A one then C. Two B. Is one. So we can go from here also. So C. Two. Let's go straight so C. Two B. Is one then see to see no relation or no urgency and see today is to C. To D. Yes. Two then D to a already done D. Two B is one so D. To be so we can go back to here with one and D. Two C zero and D. Two D. Is too again indeed two days too. So this is our solution. Thank you.

Hi. So I'm gonna be able to help you with this question. So looking at it the um it's asking the graph two X minus three Y equals 12. So the first thing you want to do is always get why? Alone to isolate why and to do that. You want to start by subtracting. So the um opposite of addition or the inverse operation of addition is subtraction. And so you want to subtract two X on this side and get two X on this side. Because um again like I said we wanted to die alone and to put we're just basically transferring um the X to the other side. So now you're left with this equation here and negative three times Y. Um Is it's still why here but it's simply saying negative three times Y. And we want to get just this white here. So um as I was saying this is kind of multiplication here. So we want the inverse operation of multiplication is division. And so you want to just divide negative three, Y by negative three. So that this cancels out. And then you want to um divide negative three on all of these here. So the new equation would be why equals 12 divided by negative three is negative four minus two arms, I plus two thirds X. Because a negative um and a negative is a positive, so negative two thirds X. And you're probably wondering well two x is two times whatever. So how come you didn't divide that? And that's because um the form we want to end up in is Y equals mx plus B. And so um if I were to divide that it would just be left with X and it wouldn't really work out. So you can even write this as because I was saying why equals mx plus B. Since we're kind of into it, we can just write it as y equals two thirds. Um Sorry what? I can see if there's x minus four and then so you want to do? So it's now to plot this, sorry, this is our final equation. So the y intercept is wherever the line crosses the graph and it's right here at negative four. So when X is zero, why is negative for that? Makes sense? And then you want to start with the graph. So if you've heard the term rise over run, I read that here rise over run. It's simply how many units to rise up and how many to go across or run across. So it says we need to um it's in this case it's two thirds. So it's saying we rise to units and then we go across three units. So it would be right here, rise up. Two units run across three. Rise up, run across. And as you can see. So even if I would go down to create that motion here, it's creating this line, this linear um, you know, consistency right now, and that's the line that you want to kind of go through. That This is a very, very not straight, not going through the line, um through the dots line, but you want to use your ruler and kind of make a line going through that. Um Thank you so much for the question. I hope you have a great rest of it.

Hello. So before we started we need to recall that. Yeah, a connected graph have no simple circuits or having no simple circuit scored a tree and the spanning tree of a simple graph. It's a subgroup of G. That is a tree containing every vertex of G. Okay, so for every versaces Of every rectus there's N -1 edges. Right? So so in the questions I suppose that you have uh in various is okay. And then um edges and of course from from here we consider then the spanning tree G. Contains anyone that's one urges. Okay, so now we're gonna so we want to produce a spanning tree by removing uh X urges. All right, So this is uh spanning tree with this number of edges. This is an edge. And we want to remove x. Number of edges from it to form and minus one. Okay, So that we can equip these two and find X. Okay, there's my algebra, so the answer will be looking for a uh minus and plus one. Okay, so there's a number of urges you're supposed to uh remove uh to produce explanatory. Okay. Thank you.


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